Decomposition of zero divisor graph into cycles and stars

A.Kuppan; J.Ravi Sankar

doi:10.1016/j.jksus.2022.102391

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Original article

01 2022

:35;

102391

doi:

10.1016/j.jksus.2022.102391

Decomposition of zero divisor graph into cycles and stars

A.Kuppan^{^a}, J.Ravi Sankar^{^a,⁎}

a Department of Mathematics, School of Advanced Sciences, Vellore Institute of Technology, Vellore, Tamil Nadu 632014, India

⁎Corresponding author. ravisankar.j@vit.ac.in (J.Ravi Sankar)

Received: 2022-7-13, Accepted: 2022-10-18,

Disclaimer:
This article was originally published by Elsevier and was migrated to Scientific Scholar after the change of Publisher.

Peer review under responsibility of Ain Shams University.

Abstract

For a graph G and a subgraph H of G, an H-decomposition of G is a partition of the edge set of G into subsets $E_{i}, 1 ⩽ i ⩽ k$ , such that each $E_{i}$ induces a graph isomorphic to H. A graph $Γ (ℝ)$ is said to be non-zero zero divisor graph of commutative ring ℝ with identity if $u, v \in V (Γ (ℝ))$ and $(u, v) \in E (Γ (ℝ))$ if and only if $uv = 0$ . It is prove that complete decomposible into cycle of length 4 of an H-decomposition of the zero divisor graph $Γ (ℝ)$ where H is any simple connected graph. In particular, we give a complete solution to the problem in the case $Z_{p} \times Z_{p} \times Z_{p} \times, \dots, \times Z_{p}$ (n times). For any positive integer $n > 2$ , there exists a decomposition of $Γ (ℝ)$ into cycle and stars in a commutative ring ℝ. We show that the obvious the graph $Γ (ℝ)$ is decomposition into cycle and stars. Overall, the proposed of the graph $Γ (ℝ)$ has significantly improved the decomposing to algebraic structure which can be useful for networking. In this paper we investigate the concept of $Γ (ℝ)$ is decomposition into cycles and stars as a commutative rings $R = Z_{p} \times Z_{p}, Z_{p} \times Z_{p} \times Z_{p}, Z_{p} \times Z_{p} \times Z_{p} \times Z_{p}$ and $Z_{p} \times Z_{p} \times Z_{p} \times, \dots, \times Z_{p}$ with p is a prime number. It is prove that the zero divisor graph $Γ (R)$ is complete decomposible into cycle of length 4 and star. In particular, we give a complete solution to the problem in the case $Z_{p} \times Z_{p} \times Z_{p} \times, \dots, \times Z_{p}$ ( $n$ times). For any positive integer $n > 2$ , there exists a decomposition of $Γ (R)$ into cycle and stars in a commutative ring $ℝ$ . We show that the obvious the graph $Γ (R)$ is decomposition into cycle and stars. Overall, the proposed of the graph $Γ (R)$ has significantly improved the decomposing to algebraic structure which can be useful for networking area.

Keywords

Graph decomposition

Zero divisor graph

Balanced complete bipartite

05C25

06A11

05C70

1 Introduction

Throughout this paper, let us take only simple, finite, non-directed graphs. For further graph theory terminology in general, refer (Bondy and Murty, 1976). Let us consider the complete graph on $p - 1$ vertices for the zero divisor graph as $Γ (Z_{p^{2}}) ≅ K_{p - 1}$ and also consider the complete bipartite graph on $p - 1$ and $q - 1$ vertices for the zero divisor graph $Γ (Z_{pq}) ≅ K_{p - 1, q - 1}$ . If R is the commutative ring and $R = Z_{p} \times Z_{p}$ then the zero divisor graph $Γ (R)$ is said to be balanced complete bipartite graph with $2 (p - 1)$ vertices. Here $C_{k}$ represents the cycle with vertices $v_{0}, v_{1}, \dots, v_{k - 1}$ and edges $v_{0} v_{1}, v_{1} v_{2}, \dots, v_{k - 2} v_{k - 1}, v_{k - 1} v_{0}$ as $(v_{0}, v_{1}, v_{2}, \dots, v_{k - 1}, v_{0})$ . Also $S_{k}$ represents the star graph with a centre vertex $v_{0}$ and k end vertices $v_{1}, v_{2}, \dots, v_{k}$ as $(v_{0}; v_{1}, v_{2}, \dots, v_{k})$ . The edges can be partitioned into $E_{1}, E_{2}, \dots, E_{k}$ such that subgraph induced by $E_{i}$ $\forall i$ , where i lies between 1 and k. Let the graph $Γ (R)$ is the decomposition of $D_{1}, D_{2}, \dots, D_{k}$ and we write $Γ (R) = D_{1} \oplus D_{2} \oplus, \dots, \oplus D_{k}$ . If $D_{i} ≅ D$ , then $Γ (R)$ is known as D-decomposition and it is denoted by $D | Γ$ where $1 ⩽ i ⩽ k, \forall i$ . If the zero divisor graph $Γ (R)$ can be decomposed into a times of $D_{1}$ and b times of $D_{2}$ , then we can say $Γ (R)$ as ${{aD}_{1}, {bD}_{2}}$ -decomposition or $(D_{1}, D_{2})$ -multidecomposition. If this decomposition is true for every a and b with necessary conditions then $Γ (R)$ has ${D_{1}, D_{2}}_{{a, b}}$ -decomposition or complete $(D_{1}, D_{2})$ -decomposition (Shyu, 2010, 2012, 2013 ).

The multidecomposition $(L_{1}, L_{2})$ was introduced by Abueida and Daven (2003) and they proved the existence of $(H_{1}, H_{2})$ -multidecomposition of $K_{m} (λ)$ when $(H_{1}, H_{2}) = (K_{1, n - 1}, C_{n})$ , where n = 3,4 and 5 (Abueida and Daven, 2004, 2007, 2000 ). Priyadharsini and Muthusamy (2009) showed the necessary and sufficient conditions for $λ K_{n}$ when $(H_{1}, H_{2})$ exists, where $H_{1}, H_{2} \in {C_{n}, P_{n}, S_{n - 1}}$ . Lee (2013) discussed the necessary and sufficient conditions for the multidecomposition of $K_{m, n}$ into minimum one copy of $C_{k}$ and $S_{k}$ . Also the necessary and sufficient conditions for the existence of decomposition of product graphs into paths and cycles with 4-edges is derived by Jeevadoss and Muthusamy (2016). Moreover Ilayaraja et.al, enterded their results for the decomposition of product of graphs into paths and stars on 5-vertices. Many other results on decomposition of zero divisor graphs into distinct subgraphs involving cycles, complete and stars have been proved in Alspach and Marshall (1994), Bryant and Maenhaut (2004) and Huang (2015).

The zero divisor graph concept was initiated by I.Beck in 1988 (Beck, 1998) and he considered zero for constructing zero divisor graphs. Few years later Anderson and Livingston, 1999 rearranged I.Beck’s definition by removing zero from his vertex set while constructing his graph. For further algebraic graph theory terminology in Kuppan and Ravi Sankar (2020)].

2 Decomposition of $Γ (Z_{p} \times Z_{p})$

Proposition 2.1

If $Γ (R)$ is a square graph, then the commutative ring $R = Z_{3} \times Z_{3}$ .

Proof

The vertices $V = {(0, 1), (0, 2), (1, 0), (2, 0)}$ are non-zero zero divisors. Therefore $Γ (R)$ is

Theorem 2.2

The graph $Γ (R)$ is a balanced complete bipartite graph iff p is any prime and $R = Z_{p} \times Z_{p}$ .

Proof

Suppose $Γ (R)$ be the balanced complete bipartite graph, then the vertex set of $V (Γ (ℝ)) = \{(u_{1}, 0), (0, u_{2}) | u_{1}, u_{2} \in {1, 2, 3, \dots, p - 1}\}$ . Now consider the vertex subsets as $V_{1} = \{(u_{1}, 0) | u_{1} \in {1, 2, 3, \dots, p - 1}\}$ and $V_{2} = \{(0, u_{2}) | u_{2} \in {1, 2, 3, \dots, p - 1}\}$ . Let $0 \neq u \in V_{1}$ or $V_{2}$ with $u^{2} \neq 0$ then no two vertices in $V_{1}$ or $V_{2}$ is non-adjacent. For any two vertices $u = (u_{1}, 0) \in V_{1}$ and $v = (0, u_{2}) \in V_{2}$ such that $uv = 0$ we say that the edges from each vertex in $V_{1}$ to every vertices in $V_{2}$ . Hence $R = Z_{p} \times Z_{p}$ . Conversely, consider the ring $Z_{p} \times Z_{p}$ with vertex set is non-zero zero divisors $V = \{(0, u_{2}), (u_{1}, 0) | u_{1}, u_{2} \in {1, 2, 3, \dots, p - 1}\}$ . For all $u \in V_{1}, v \in V_{2}$ with $u^{2} \neq 0$ and $v^{2} \neq 0$ . Suppose $uv = 0$ there exists an edge from u to v. Clearly, the commutative ring $R = Z_{p} \times Z_{p}$ is a graph of $Γ (R)$ and it is a balanced complete bipartite graph.

Theorem 2.3

Let p be any odd prime then the graph $Γ (Z_{p} \times Z_{p})$ is decomposition into $\frac{{(p - 1)}^{2}}{4}$ copies of $C_{4}$ .

Proof

The vertex set of $Γ (Z_{p} \times Z_{p}) = \{(u_{1}, 0), (0, u_{2})\}$ such that $u_{1}, u_{2}$ lies between 1 and $p - 1$ . That is $| V (Γ (Z_{p} \times Z_{p})) | = 2 (p - 1)$ . Let $V_{1}$ and $V_{2}$ be the partition of vertex subsets where $V_{1} = \{(u_{1}, 0) | u_{1} \in {0, 1, 2, \dots, p - 1}\}$ and $V_{2} = \{(0, u_{2}) | u_{2} \in {0, 1, 2, \dots, p - 1}\}$ . That is $| V_{1} | = p - 1$ and $| V_{2} | = p - 1$ . Let any two vertices $u, v \in V_{1}$ or $V_{2}$ then $uv \neq 0$ . Clearly u and v are disconnected.

Let us consider the any one member $u = (u_{1}, 0)$ in $V_{1}$ and another member $v = (0, u_{2})$ in $V_{2}$ then $(u_{1}, 0) . (0, u_{2}) = (0, 0)$ . Clearly u and v are connected.

Therefore, each vertex of $V_{1}$ is connected to every vertices in $V_{2}$ . That is $Γ (Z_{p} \times Z_{p})$ is balanced complete bipartite graph in the form of $K_{p - 1, p - 1}$ .Clearly, $Γ (Z_{p} \times Z_{p})$ can be decomposed into $\frac{{(p - 1)}^{2}}{4}$ copies of $C_{4}$ . That is $\begin{matrix} ((0, 1), (1, 0), (0, p - 1), (p - 1, 0), (0, 1)), ((0, 1), (2, 0), (0, p - 1), (p - 2, 0), (0, 1)), \dots, ((0, 1), (⌊ \frac{p}{2} ⌋, 0), (0, p - 1), (⌊ \frac{p}{2} ⌋, 0), (0, 1)), \\ ((0, 2), (1, 0), (0, p - 2), (p - 1, 0), (0, 2)), ((0, 2), (2, 0), (0, p - 2), (p - 2, 0), (0, 2)), \dots, ((0, 2), (⌊ \frac{p}{2} ⌋, 0), (0, p - 2), (⌊ \frac{p}{2} ⌋, 0), (0, 2)), \\ ((0, 3), (1, 0), (0, p - 3), (p - 1, 0), (0, 3)), ((0, 3), (2, 0), (0, p - 3), (p - 2, 0), (0, 3)), \dots, ((0, 3), (⌊ \frac{p}{2} ⌋, 0), (0, p - 3), (⌊ \frac{p}{2} ⌋, 0), (0, 3)) \\ ⋮ \\ ((0, ⌊ \frac{p}{2} ⌋), (⌊ \frac{p}{2} ⌋, 0), (0, ⌊ \frac{p}{2} ⌋ + 1), (⌊ \frac{p}{2} ⌋ + 1, 0), (0, ⌊ \frac{p}{2} ⌋)) \end{matrix}$ total values of each cycle is $\frac{{(p - 1)}^{2}}{4}$ . Therefore, $Γ (Z_{p} \times Z_{p})$ can be decomposed into $\frac{{(p - 1)}^{2}}{4}$ copies of $C_{4}$ .

Example 1

Let us consider $p = 11$ and the vertex set of $Γ (Z_{11} \times Z_{11})$ as

$V = \{(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (0, 10), (1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (7, 0), (8, 0), (9, 0), (10, 0)\}$ .

Then $V (Γ (Z_{11} \times Z_{11}))$ can be split into two parts, namely $V_{1}$ and $V_{2}$ .

$V_{1} = \{(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (0, 10)\}$ and.

$V_{2} = \{(1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (7, 0), (8, 0), (9, 0), (10, 0)\}$ .

Clearly, $Γ (Z_{11} \times Z_{11})$ is isomorphic to $K_{10, 10}$ . The Figs. 1,2 indicates the decomposition of the graph $K_{10, 10}$ into 25 copies of $C_{4}$ .

Theorem 2.4

Let p be odd prime and $a, b$ are the non-negative integers. If there exists $(C_{4}, S_{4})$ -decomposition of $Γ (Z_{p} \times Z_{p})$ , then ${(p - 1)}^{2} \equiv 0 (\mod 4)$ .

Theorem 2.5

Let p be odd prime, the graph $Γ (Z_{p} \times Z_{p})$ is $S_{4}$ -decomposable. Then $If p - 1 \equiv 0 (\mod 4) then S_{4} is decomposable$ $If p - 1 ≢ 0 (\mod 4) then S_{4} is non - decomposable$

Proof

We shall prove this theorem by using the following two cases.

Case(i): Let $p - 1 \equiv 0 (\mod 4)$ . Assume that $p - 1$ is divided by 4. Let us consider the vertex set of non-zero zero-divisors $V (Γ (Z_{p} \times Z_{p})) = \{(0, x), (x, 0) | x \in {0, 1, 2, \dots, p - 1}\}$ . Here $Γ (Z_{p} \times Z_{p})$ be the zero divisor graph of a ring $Z_{p} \times Z_{p}$ . If there is a member $0 \neq (0, x) \in V$ then $(0, x), (0, x) \neq 0$ such that the two zero divisors are non-adjacent. If there is a member $0 \neq (0, x) \in V$ and $0 \neq (x, 0) \in V$ then, $(0, x), (x, 0) = 0 \in V$ such that two zero divisors are adjacent. That is $| V (Γ (Z_{p} \times Z_{p})) | = 2 (p - 1)$ . Clearly, the graph completely decomposed star with size 4.Case(ii): Let $p - 1 ≢ 0 (\mod 4)$ . Assume that $p - 1$ is not divided by 4. If there is a member $0 \neq (0, x) \in V$ then $(0, x), (0, x) \neq 0$ such that two zero divisors are non-adjacent. Clearly, the vertex $(0, x) \in V$ is not completely decomposed into the star graph of size 4.

Lemma 2.6

If a and b are any two positive integers, then there exists a complete ${C_{4}, S_{4}}$ -decomposition of $Γ (Z_{5} \times Z_{5})$ .

Proof

Let $V (Γ (Z_{5} \times Z_{5})) = \{(0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (2, 0), (3, 0), (4, 0)\}$ . Then the required entire decomposition of ${C_{4}, S_{4}}$ are

$a = 4$ and $b = 0$ . The required cycles are

$((0, 1), (1, 0), (0, 4), (4, 0), (0, 1)), ((0, 1), (2, 0), (0, 4), (3, 0), (0, 1)), ((0, 2), (1, 0), (0, 3), (4, 0), (0, 2)), ((0, 2), (2, 0), (0, 3), (3, 0), (0, 2))$ .
$a = 2$ and $b = 2$ . The required cycles and stars are

$((0, 1), (1, 0), (0, 4), (4, 0), (0, 1)), ((0, 1), (2, 0), (0, 4), (3, 0), (0, 1)), ((0, 2), (1, 0), (2, 0), (3, 0), (4, 0)), ((0, 3), (1, 0), (2, 0), (3, 0), (4, 0))$ .
$a = 0$ and $b = 4$ . The required stars are $((0, 1), (1, 0), (2, 0), (3, 0), (4, 0)), ((0, 2), (1, 0), (2, 0), (3, 0), (4, 0)), ((0, 3), (1, 0), (2, 0), (3, 0), (4, 0)), ((0, 4), (1, 0), (2, 0), (3, 0), (4, 0))$ .

Lemma 2.7

If a and b are any two positive integers, then there exists a complete ${C_{4}, S_{6}}$ -decomposition of $Γ (Z_{7} \times Z_{7})$ .

Proof

Let $V (Γ (Z_{7} \times Z_{7})) = \{(1, 0), (2, 0), (3, 0), (4, 0), (5, 0), (6, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6)\}$ . Then the required entire decomposition of ${C_{4}, S_{6}}$ are

$a = 9$ and $b = 0$ the required cycles are

$((0, 1), (1, 0), (0, 6), (6, 0), (0, 1)), ((0, 1), (2, 0), (0, 6), (5, 0), (0, 1)), ((0, 1), (3, 0), (0, 6), (4, 0), (0, 1)),$ $((0, 2), (1, 0), (0, 5), (6, 0), (0, 2)), ((0, 2), (2, 0), (0, 5), (5, 0), (0, 2)), ((0, 2), (3, 0), (0, 5), (4, 0), (0, 2))$ ,

$((0, 3), (1, 0), (0, 4), (6, 0), (0, 3)), ((0, 3), (2, 0), (0, 4), (5, 0), (0, 3)), ((0, 3), (3, 0), (0, 4), (4, 0), (0, 3))$ .
$a = 6$ and $b = 2$ the required cycles and stars are

$((0, 1), (1, 0), (0, 6), (6, 0), (0, 1)), ((0, 1), (2, 0), (0, 6), (5, 0), (0, 1)), ((0, 1), (3, 0), (0, 6), (4, 0), (0, 1)),$ $((0, 2), (1, 0), (0, 5), (6, 0), (0, 2)), ((0, 2), (2, 0), (0, 5), (5, 0), (0, 2)), ((0, 2), (3, 0), (0, 5), (4, 0), (0, 2))$ ,

$((0, 3); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 4); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ .
$a = 3$ and $b = 4$ the required cycles and stars are

$((0, 1), (1, 0), (0, 6), (6, 0), (0, 1)), ((0, 1), (2, 0), (0, 6), (5, 0), (0, 1)), ((0, 1), (3, 0), (0, 6), (4, 0), (0, 1)),$ $((0, 2); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 3); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 4); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 5); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ .
$a = 0$ and $b = 6$ the required cycles and stars are

$((0, 1); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 2); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 3); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 4); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ ,

$((0, 5); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ , $((0, 6); (1, 0), (2, 0), (3, 0), (4), (5, 0), (6, 0))$ .

Theorem 2.8

Let a and b be any two positive integers and p is any prime number, then there exists a complete $(C_{4}, S_{4})$ -decomposition of $Γ (Z_{p} \times Z_{p})$ for all $p ⩾ 5$ .

Proof

If $p = 5, 7$ , then the result follows by Lemmas 2.6 and 2.7. For $p > 7$ , we write, $Γ (Z_{p} \times Z_{p}) = \{(u_{i}, 0), (0, u_{i})\}$ where $u_{i}$ from 1 to $p - 1$ . Let $p - 1 \equiv 0 (\mod 4)$ . By Lemma 2.6 and 2.7, the graphs $K_{4, 4}, K_{6, 6}$ have a complete ${C_{4}, S_{4}}$ -decomposition. Clearly, the graph $Γ (Z_{p} \times Z_{p})$ has the desired decomposition.

3 Decomposition of $Γ (Z_{p} \times Z_{p} \times Z_{p})$

Theorem 3.1

For any odd prime p then $Γ (Z_{p} \times Z_{p} \times Z_{p})$ is decomposition into 3 copies of $K_{{(p - 1)}^{2}, p - 1}$ and 3 copies of $K_{p - 1, p - 1}$ .

Proof

Let $Γ (Z_{p} \times Z_{p} \times Z_{p})$ be the zero divisor graph of a commutative ring $Z_{p} \times Z_{p} \times Z_{p}$ . The vertex set $V = \{(a, 0, 0), (0, b, 0), (0, 0, c), (a, b, 0), (a, 0, c), (0, b, c) | a, b, c \in {1, 2, 3, \dots, p - 1\}}$ where p is any odd prime. The vertex set can be partitioned into several disjoint subsets $V_{1} = \{(a, 0, 0) | a \in {1, 2, 3, \dots, p - 1}\}, V_{2} = \{(0, b, 0) | b \in {1, 2, 3, \dots, p - 1}\}, V_{3} = \{(0, 0, c) | c \in {1, 2, 3, \dots, p - 1}\}, V_{4} = \{(a, b, 0) | a, b \in {1, 2, 3, \dots, p - 1}\}, V_{5} = \{(a, 0, c) | a, c \in {1, 2, 3, \dots, p - 1}\}$ and $V_{6} = \{(0, b, c) | b, c \in {1, 2, 3, \dots, p - 1}\}$ . That is $| V_{1} | = | V_{2} | = | V_{3} | = p - 1$ and $| V_{4} | = | V_{5} | = | V_{6} | = {(p - 1)}^{2}$ .

Case(i): Consider the vertex subsets $V_{1}, V_{2}$ and $V_{3}$ . If $u \in$ $V_{1}$ with $u^{2} \neq 0$ then the vertices are non-adjacent. If $v \in$ $V_{2}$ with $uv = 0$ then there is an edge between every pair of vertices. Clearly, the above subsets are adjacent to each other. By continuing the same process we can prove the same with $(V_{2}, V_{3})$ and $(V_{1}, V_{3})$ . Hence, 3 copies of $K_{p - 1, p - 1}$ .

Case(ii): Consider the vertex subsets $V_{4}, V_{5}$ and $V_{6}$ . If $u \in$ $V_{4}$ with $u^{2} \neq 0$ then the vertices are non-adjacent. If $v \in$ $V_{3}$ with $uv = 0$ then there is an edge between every pair of vertices. Clearly, the above subsets are adjacent to each other. Continuing the same process we can prove the same with $(V_{1}, V_{6})$ and $(V_{2}, V_{5})$ . Hence, 3 copies of $K_{p - 1, {(p - 1)}^{2}}$ . Therefore, from the above two cases, indicate the graph $Γ (Z_{p} \times Z_{p} \times Z_{p})$ is decomposition into 3 copies of $K_{p - 1, p - 1}$ and 3 copies of $K_{p - 1, {(p - 1)}^{2}}$ .

Theorem 3.2

For any odd prime p, then $Γ (Z_{p} \times Z_{p} \times Z_{p})$ is decomposed into $\frac{3}{4} ({(p - 1)}^{3} + {(p - 1)}^{2})$ copies of $C_{4}$ .

Proof

If p is an odd prime, then the vertices are non-zero zero divisors $V (Γ (Z_{p} \times Z_{p} \times Z_{p})) = \{(a_{i}, 0, 0), (0, b_{i}, 0), (0, 0, c_{i}), (a_{i}, b_{i}, 0), (a_{i}, 0, c_{i}), (0, b_{i}, c_{i})\}$ where $1 ⩽ a_{i}, b_{i}, c_{i} ⩽ p - 1$ . That is $| V (Γ (Z_{p} \times Z_{p} \times Z_{p})) | = 3 (p - 1) + 3 {(p - 1)}^{2}$ . Let us take the vertex subsets are $V_{1} = \{(a_{i}, 0, 0) | 1 ⩽ a_{i} ⩽ p - 1}, V_{2} = {(0, b_{i}, 0) | 1 ⩽ b_{i} ⩽ p - 1\}, V_{3} = \{(0, 0, c_{i}) | 1 ⩽ c_{i} ⩽ p - 1\}, V_{4} = \{(a_{i}, b_{i}, 0) | 1 ⩽ a_{i}, b_{i} ⩽ p - 1\}, V_{5} = \{(a_{i}, 0, c_{i}) | 1 ⩽ a_{i}, c_{i} ⩽ p - 1\}$ and $V_{6} = \{(0, b_{i}, c_{i}) | 1 ⩽ b_{i}, c_{i} ⩽ p - 1\}$ . That is $| V_{1} | = | V_{2} | = | V_{3} | = p - 1$ and $| V_{4} | = | V_{5} | = | V_{6} | = {(p - 1)}^{2}$ . Let us prove this by the following two cases.

Case(i): Let us take the pairs of vertex subsets $(V_{1}, V_{6}), (V_{2}, V_{5})$ and $(V_{3}, V_{4})$ . Here $(V_{1}, V_{6}) ≅ K_{p - 1, {(p - 1)}^{2}}$ . Similarly we say that $(V_{2}, V_{5}) ≅ K_{p - 1, {(p - 1)}^{2}}$ and $(V_{3}, V_{4}) ≅ K_{p - 1, {(p - 1)}^{2}}$ .

Case(ii): Let us take the vertex subsets are $V_{1}, V_{2}$ and $V_{3}$ . Then every pair of above vertex subsets are isomorphic to tri-partite graph of $K_{p - 1, p - 1, p - 1}$ . $\begin{matrix} D (K_{p - 1, p - 1}^{V_{1}, V_{2}}) = \\ ((a_{1}, 0, 0), (0, b_{1}, 0), (a_{p - 1}, 0, 0), (0, b_{p - 1}, 0), (a_{1}, 0, 0)), \\ ((a_{1}, 0, 0), (0, b_{2}, 0), (a_{p - 1}, 0, 0), (0, b_{p - 2}, 0), (a_{1}, 0, 0)), \\ ((a_{1}, 0, 0), (0, b_{3}, 0), (a_{p - 1}, 0, 0), (0, b_{p - 3}, 0), (a_{1}, 0, 0)), \\ ((a_{2}, 0, 0), (0, b_{1}, 0), (a_{p - 2}, 0, 0), (0, b_{p - 1}, 0), (a_{2}, 0, 0)), \\ ((a_{2}, 0, 0), (0, b_{2}, 0), (a_{p - 2}, 0, 0), (0, b_{p - 2}, 0), (a_{2}, 0, 0)), \\ ((a_{2}, 0, 0), (0, b_{3}, 0), (a_{p - 2}, 0, 0), (0, b_{p - 3}, 0), (a_{2}, 0, 0)), \\ ((a_{\frac{p - 1}{2}}, 0, 0), (0, b_{\frac{p - 1}{2}}, 0), (a_{\frac{p - 1}{2} + 1}, 0, 0), (0, b_{\frac{p - 1}{2} + 1}, 0), (a_{\frac{p - 1}{2}}, 0, 0)) . \end{matrix}$

On the whole there are ${(\frac{p - 1}{2})}^{2}$ copies of $C_{4}$ . If we continue the same process we get ${(\frac{p - 1}{2})}^{2}$ copies for $D (K_{p - 1, p - 1}^{V_{1}, V_{3}})$ and $D (K_{p - 1, p - 1}^{V_{2}, V_{3}})$ . Clearly, from the above cases we get $\frac{3}{4} ({(p - 1)}^{3} + {(p - 1)}^{2})$ copies of $C_{4}$ . Therefore, $Γ (Z_{p} \times Z_{p} \times Z_{p})$ can be decomposed into $\frac{3}{4} ({(p - 1)}^{3} + {(p - 1)}^{2})$ copies of $C_{4}$ .

Theorem 3.3

Let p be any odd prime, then $Γ (Z_{p} \times Z_{p} \times Z_{p})$ is decomposed into ${(p - 1)}^{2}$ copies of $S_{p - 1}$ and $\frac{3 {(p - 1)}^{2}}{4}$ copies of $C_{4}$ .

Proof

Consider the vertex set of $Γ (Z_{p} \times Z_{p} \times Z_{p})$ is $\{(x, 0, 0), (0, y, 0), (0, 0, z), (0, y, z), (x, 0, z), (x, y, 0) | x, y, z \in {1, 2, 3, \dots, p - 1}\}$ . Let us consider the following subsets $V_{1}, V_{2}, V_{3}, V_{4}, V_{5}$ and $V_{6}$ are in $V (Γ (Z_{p} \times Z_{p} \times Z_{p}))$ .

Case(i): Let us take the vertex subsets $V_{1}, V_{2}$ and $V_{3}$ . Then every pair of subsets are isomorphic to $K_{p - 1, p - 1}$ . Cardinality of above all subsets is $p - 1$ . If $x \in V_{1}$ or $V_{2}$ or $V_{3}$ and $x^{2} \neq 0$ then they are non adjacent to each other. If every distinct pair of vertices are connected then there exists a tripartite graph. If the size of the pair $(V_{1}, V_{2})$ is ${(p - 1)}^{2}$ then the decomposition of the graph $K_{p - 1, p - 1}$ yields $\frac{{(p - 1)}^{2}}{4}$ copies. Similarly we can say that, if the sizes of $(V_{2}, V_{3})$ and $(V_{1}, V_{3})$ is ${(p - 1)}^{2}$ then the decomposition of the graph $K_{p - 1, p - 1}$ has $\frac{{(p - 1)}^{2}}{4}$ copies.Case(ii): Let us take the pairs of vertex subsets $(V_{1}, V_{4}), (V_{2}, V_{5})$ and $(V_{3}, V_{6})$ . Then every pair of subsets are isomorphic to $K_{p - 1, {(p - 1)}^{2}}$ . Cardinality of above all pairs of subsets is $(p - 1) + {(p - 1)}^{2}$ . If $x \in V$ and $x^{2} \neq 0$ then they are non adjacent to each other. If the size of the pair $(V_{1}, V_{6})$ is ${(p - 1)}^{3}$ then the decomposition of the graph $K_{p - 1, {(p - 1)}^{2}}$ yields $\frac{{(p - 1)}^{3}}{4}$ copies. Similarly we can say that, if the sizes of $(V_{2}, V_{5})$ and $(V_{3}, V_{4})$ is ${(p - 1)}^{3}$ then the decomposition of the graph $K_{p - 1, {(p - 1)}^{2}}$ has $\frac{{(p - 1)}^{3}}{4}$ copies. Clearly, from the above cases, we get ${(p - 1)}^{2}$ copies of $S_{p - 1}$ and $\frac{3 {(p - 1)}^{2}}{4}$ copies of $C_{4}$ .

Theorem 3.4

Let p be any odd prime, then $Γ (Z_{p} \times Z_{p} \times Z_{p})$ can be decomposed into $(\binom{p - 1}{2})$ copies of $C_{6}, p - 1$ copies of $C_{3}$ and $\frac{3 {(p - 1)}^{3}}{4}$ copies of $C_{4}$ .

Proof

Let the vertex set of non-zero zero divisors is $V (Γ (Z_{p} \times Z_{p} \times Z_{p})) = \{(a_{i}, 0, 0), (0, b_{i}, 0), (0, 0, c_{i}), (a_{i}, b_{i}, 0), (a_{i}, 0, c_{i}), (0, b_{i}, c_{i})\}$ where $1 ⩽ a_{i}, b_{i}, c_{i} ⩽ p - 1$ . That is $| V (Γ (Z_{p} \times Z_{p} \times Z_{p})) | = 3 (p - 1) + 3 {(p - 1)}^{2}$ . The partition of vertex subsets are $V_{1} = \{(a_{i}, 0, 0) | 1 ⩽ a_{i} ⩽ p - 1\} = \{u_{1}, u_{2}, u_{3}, \dots, u_{p - 1}\}, V_{2} = \{(0, b_{i}, 0) | 1 ⩽ b_{i} ⩽ p - 1\} = \{v_{1}, v_{2}, v_{3}, \dots, v_{p - 1}\}, V_{3} = \{(0, 0, c_{i}) | 1 ⩽ c_{i} ⩽ p - 1\} = \{w_{1}, w_{2}, w_{3}, \dots, w_{p - 1}\}, V_{4} = \{(a_{i}, b_{i}, 0) | 1 ⩽ a_{i}, b_{i} ⩽ p - 1\}, V_{5} = \{(a_{i}, 0, c_{i}) | 1 ⩽ a_{i}, c_{i} ⩽ p - 1\}$ and $V_{6} = \{(0, b_{i}, c_{i}) | 1 ⩽ b_{i}, c_{i} ⩽ p - 1\}$ . That is $| V_{1} | = | V_{2} | = | V_{3} | = p - 1$ and $| V_{4} | = | V_{5} | = | V_{6} | = {(p - 1)}^{2}$ . We can prove this proof by the following cases.

Case(i): Consider the vertex subsets $V_{1}, V_{2}$ and $V_{3}$ . If we decompose the tripartite graph then we get $(\binom{p - 1}{2})$ copies of $C_{6}$ and $p - 1$ copies of $C_{3}$ . Which are written as follows. $\begin{matrix} (u_{1}, v_{2}, w_{1}, u_{2}, v_{1}, w_{2}, u_{1}), (u_{1}, v_{3}, w_{1}, u_{3}, v_{1}, w_{3}, u_{1}), \\ \dots, (u_{1}, v_{p - 2}, w_{1}, u_{p - 2}, v_{1}, w_{p - 2}, u_{1}) . \\ (u_{2}, v_{3}, w_{2}, u_{3}, v_{2}, w_{3}, u_{2}), (u_{2}, v_{4}, w_{2}, u_{4}, v_{2}, w_{4}, u_{2}), \\ \dots, (u_{2}, v_{p - 2}, w_{2}, u_{p - 2}, v_{2}, w_{p - 2}, u_{2}) . \\ ⋮ \\ (u_{p - 3}, v_{p - 2}, w_{p - 3}, u_{p - 2}, v_{p - 3}, w_{p - 2}, u_{p - 3}) . \end{matrix}$

If we remove $(\binom{p - 1}{2})$ copies of $C_{6}$ then the resultant graph will have $(u_{1}, v_{1}, w_{1}), (u_{2}, v_{2}, w_{2})$ and $(u_{3}, v_{3}, w_{3})$ copies of $C_{3}$ edges only. Here the degree of each vertex is of even degree.

Case(ii): Consider the following vertex subsets $V_{4}, V_{5}$ and $V_{6}$ . That is $| V_{4} | = | V_{5} | = | V_{6} | = {(p - 1)}^{2}$ . By Theorem 3.3 we say that $\frac{3 {(p - 1)}^{3}}{4}$ copies of $C_{4}$ . Therefore, from the above two cases, we get $Γ (Z_{p} \times Z_{p} \times Z_{p})$ is decomposition into $(\binom{p - 1}{2})$ copies of $C_{6}, p - 1$ copies of $C_{3}$ and $\frac{3 {(p - 1)}^{3}}{4}$ copies of $C_{4}$ .

4 Decomposition of $Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p})$

Theorem 4.1

If p is any odd prime, then $Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p})$ is decomposed into 3-copies of $K_{{(p - 1)}^{2}, {(p - 1)}^{2}}$ , 4-copies of $K_{{(p - 1)}^{3}, (p - 1)}$ , 6-copies of $K_{(p - 1), (p - 1)}$ and 12-copies of $K_{{(p - 1)}^{2}, (p - 1)}$ .

Proof

Let the vertex set $V (Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p}))$ is $V = \{(k, 0, 0, 0), (0, l, 0, 0), (0, 0, m, 0), (0, 0, 0, n), (k, l, 0, 0), (k, 0, m, 0), (k, 0, 0, n), (0, l, m, 0), (0, l, 0, n), (0, 0, m, n), (k, l, m, 0), (k, 0, m, n), (k, l, 0, n), (0, l, m, n) | k, l, m, n \in {1, 2, 3, \dots, p - 1}\}$ . The vertex set can be split into disjoint subsets, such as $A_{i}, B_{i}, C_{i} \in V$ , where $A_{1} = \{(k, 0, 0, 0) | 1 ⩽ k ⩽ p - 1\}, A_{2} = \{(0, l, 0, 0) | 1 ⩽ l ⩽ p - 1\}, A_{3} = \{(0, 0, m, 0) | 1 ⩽ m ⩽ p - 1\}, A_{4} = \{(0, 0, 0, n) | 1 ⩽ n ⩽ p - 1\}, B_{1} = \{(k, l, 0, 0) | 1 ⩽ k, l ⩽ p - 1\}, B_{2} = \{(k, 0, m, 0) | 1 ⩽ k, m ⩽ p - 1\}, B_{3} = \{(k, 0, 0, n) | 1 ⩽ k, n ⩽ p - 1\}, B_{4} = \{(0, l, m, 0) | 1 ⩽ l, m ⩽ p - 1\}, B_{5} = \{(0, l, 0, n) | 1 ⩽ l, n ⩽ p - 1\}, B_{6} = \{(0, 0, m, n) | 1 ⩽ m, n ⩽ p - 1\}, C_{1} = \{(k, l, m, 0) | 1 ⩽ k, l, m ⩽ p - 1\}, C_{2} = \{(k, 0, m, n) | 1 ⩽ k, m, n ⩽ p - 1\}, C_{3} = \{(k, l, 0, n) | 1 ⩽ k, l, n ⩽ p - 1\}, C_{4} = \{(0, l, m, n) | 1 ⩽ l, m, n ⩽ p - 1\}$ . That is $| \cup_{i = 1}^{p - 1 C_{1}} A_{i} | = 4 (p - 1), | \cup_{i = 1}^{p - 1 C_{2}} B_{i} | = 6 {(p - 1)}^{2}$ and $| \cup_{i = 1}^{p - 1 C_{3}} C_{i} | = 4 {(p - 1)}^{3}$ .

Case(i): Consider the vertex subsets $A_{i}$ and $C_{i}$ . If $u \in$ $C_{i}$ with $u^{2} \neq 0$ then the vertices are non-adjacent. If $v \in$ $A_{i}$ with $uv = 0$ then there is an edge between every pairs of vertices. Clearly, the above subsets are adjacent to each other. Clearly, we get 4 copies of $K_{p - 1, {(p - 1)}^{2}}$ .

Case(ii): Consider the vertex subsets $A_{i}$ and $B_{i}$ . If $u \in$ $B_{i}$ with $u^{2} \neq 0$ then the vertices are non-adjacent. If $v \in$ $A_{i}$ with $uv = 0$ then every pair of vertices are connected. Clearly, the above subsets $(A_{i}, B_{i})$ are isomorphic to $K_{p - 1, {(p - 1)}^{2}}$ . Therefore, we get 4 copies of $K_{{(p - 1)}^{3}, p - 1}$ .

Case(iii): Consider the vertex subset of $B_{i}$ . If $u \in$ $B_{i}$ with $u^{2} = 0$ then the vertices are adjacent. Clearly, every pair of $B_{i}$ is adjacent to itself. Therefore, we get 3 copies of $K_{{(p - 1)}^{2}, {(p - 1)}^{2}}$ .

Case(iv): Consider the vertex subset $A_{i}$ . If $u \in$ $A_{i}$ with $u^{2} = 0$ then the vertices are adjacent. Clearly, the every pair of $A_{i}$ is adjacent to itself. Therefore, we get 6 copies of $K_{(p - 1), (p - 1)}$ .

From the above four cases we say that $Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p})$ is decomposed into 3-copies of $K_{{(p - 1)}^{2}, {(p - 1)}^{2}}$ , 4-copies of $K_{{(p - 1)}^{3}, (p - 1)}$ , 6-copies of $K_{(p - 1), (p - 1)}$ and 12-copies of $K_{{(p - 1)}^{2}, (p - 1)}$ .

Theorem 4.2

If p is any odd prime, then $Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p})$ is decomposed into $p - 1$ copies of $K_{4}$ and $\frac{7 {(p - 1)}^{4} + 12 {(p - 1)}^{3} + 6 {(p - 1)}^{2}}{4}$ copies of $C_{4}$ .

Proof

The vertex set of $V (Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p})) = \{(k, 0, 0, 0), (0, l, 0, 0), (0, 0, m, 0), (0, 0, 0, n), (k, l, 0, 0), (k, 0, m, 0), (k, 0, 0, n), (0, l, m, 0), (0, l, 0, n), (0, 0, m, n), (k, l, m, 0), (k, 0, m, d), (k, l, 0, d), (0, l, m, n) | k, l, m, n \in {1, 2, 3, \dots, p - 1}\}$ . The vertex set can be split into disjoint subsets, such as $A_{i}, B_{i}, C_{i} \in V$ , where $A_{1} = \{(k, 0, 0, 0) | 1 ⩽ k ⩽ p - 1\}, A_{2} = \{(0, l, 0, 0) | 1 ⩽ l ⩽ p - 1\}, A_{3} = \{(0, 0, m, 0) | 1 ⩽ m ⩽ p - 1\}, A_{4} = \{(0, 0, 0, n) | 1 ⩽ n ⩽ p - 1\}, B_{1} = \{(k, l, 0, 0) | 1 ⩽ k, l ⩽ p - 1\}, B_{2} = \{(k, 0, m, 0) | 1 ⩽ k, m ⩽ p - 1\}, B_{3} = \{(k, 0, 0, n) | 1 ⩽ k, n ⩽ p - 1\}, B_{4} = \{(0, l, m, 0) | 1 ⩽ l, m ⩽ p - 1\}, B_{5} = \{(0, l, 0, n) | 1 ⩽ l, n ⩽ p - 1\}, B_{6} = \{(0, 0, m, n) | 1 ⩽ m, n ⩽ p - 1\}, C_{1} = \{(k, l, m, 0) | 1 ⩽ k, l, m ⩽ p - 1\}, C_{2} = \{(k, 0, m, n) | 1 ⩽ k, m, n ⩽ p - 1\}, C_{3} = \{(k, l, 0, n) | 1 ⩽ k, l, n ⩽ p - 1\}, C_{4} = \{(0, l, m, n) | 1 ⩽ l, m, n ⩽ p - 1\}$ . That is $| \cup_{i = 1}^{p - 1 C_{1}} A_{i} | = 4 (p - 1), | \cup_{i = 1}^{p - 1 C_{2}} B_{i} | = 6 {(p - 1)}^{2}$ and $| \cup_{i = 1}^{p - 1 C_{3}} C_{i} | = 4 {(p - 1)}^{3}$ .

Case(i): Let us consider the vertex subsets.

$A_{i} = \{(k, 0, 0, 0), (0, l, 0, 0), (0, 0, m, 0), (0, 0, 0, n) | k, l, m, n \in {1, 2, 3, \dots, p - 1}\}$ . If $u, v \in A_{i}$ with $uv = 0$ then there is an edge between every pair of vertices. Clearly, we get 4 copies of complete graph $K_{p - 1}$ .

Case(ii): Let us consider the pair of subsets $(A_{i}, C_{i}), (A_{i}, B_{i})$ and $(B_{i}, B_{i})$ (using Theorem 4.1). The resultant of decomposition of $(A_{i}, C_{i})$ is isomorphic to $K_{p - 1, {(p - 1)}^{3}}$ . Similarly the resultant of decomposition of $(A_{i}, B_{i})$ and $(B_{i}, B_{i})$ is isomorphic to $K_{p - 1, {(p - 1)}^{2}}$ and $K_{{(p - 1)}^{2}, {(p - 1)}^{2}}$ . Then we get cycles of length 4 and described as follows. $\begin{matrix} D (4 K_{p - 1, {(p - 1)}^{3}}) = \frac{4 (p - 1) {(p - 1)}^{3}}{4} \\ D (12 K_{p - 1, {(p - 1)}^{2}}) = \frac{12 (p - 1) {(p - 1)}^{2}}{4} \\ D (6 K_{p - 1, p - 1}) = \frac{6 (p - 1) (p - 1)}{4} \\ D (3 K_{{(p - 1)}^{2}, {(p - 1)}^{2}}) = \frac{3 {(p - 1)}^{2} {(p - 1)}^{2}}{4} \end{matrix}$

Sum of above decomposition $= \frac{4 (p - 1) {(p - 1)}^{3}}{4} + \frac{12 (p - 1) {(p - 1)}^{2}}{4} + \frac{6 (p - 1) (p - 1)}{4} + \frac{3 {(p - 1)}^{2} {(p - 1)}^{2}}{4}$ .

Therefore, from the above two cases we get $Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p})$ is decomposed into $p - 1$ copies of $K_{4}$ and $\frac{7 {(p - 1)}^{4} + 12 {(p - 1)}^{3} + 6 {(p - 1)}^{2}}{4}$ copies of $C_{4}$ .

5 Decomposition of $Γ (Z_{p} \times Z_{p} \times Z_{p} \times \dots \times Z_{p})$

Theorem 5.1

If p is odd prime and m is any non-negative integer, then $Γ (Z_{p} \times Z_{p} \times Z_{p} \times \dots \times Z_{p})$ has $m - 1$ partitions.

Proof

The vertex set of $\begin{matrix} Γ (Z_{p} \times Z_{p} \times Z_{p} \times \dots \times Z_{p}) = \\ \{((u_{1}, 0, 0, \dots, 0), (0, u_{2}, 0, \dots, 0), (0, 0, u_{3}, \dots, 0), \dots, (0, 0, 0, \dots, u_{m})), \\ ((u_{1}, u_{2}, 0, \dots, 0), (u_{1}, 0, u_{3}, \dots, 0), \dots, (upto (p - 1) (\frac{m}{2}))), \\ ((u_{1}, u_{2}, u_{3}, \dots, 0), \dots ., (upto (p - 1) (\frac{m}{3}))), \\ \dots, ((u_{1}, u_{2}, u_{3}, \dots, u_{m - 1}, 0), \dots ., (upto (p - 1) (\frac{m}{m - 1})))\} & . \end{matrix}$

The vertex set can be split into.

$V_{1} = {(u_{1}, 0, 0, \dots, 0), (0, u_{2}, 0, \dots, 0), (0, 0, u_{3}, \dots, 0), \dots, (0, 0, 0, \dots, u_{m})}$ ,

$V_{2} = {(u_{1}, u_{2}, 0, \dots, 0), (u_{1}, 0, u_{3}, \dots, 0), \dots, (upto (p - 1) (\frac{m}{2}))}$ ,

$V_{3} = {(u_{1}, u_{2}, u_{3}, \dots, 0), \dots ., (upto (p - 1) (\frac{m}{3}))}$ ⋮

$V_{m - 1} = {(u_{1}, u_{2}, u_{3}, \dots, u_{m - 1}, 0), \dots ., (upto (p - 1) (\frac{m}{m - 1}))}$ where $Z_{p} \times Z_{p} \times Z_{p} \times \dots \times Z_{p}$ (up to m times). That is $| V_{1} | = (p - 1) (\frac{m}{1}), | V_{2} | = (p - 1) (\frac{m}{2}), | V_{3} | = (p - 1) (\frac{m}{3})$ , similarly $| V_{m - 1} | = (p - 1) (\frac{m}{m - 1})$ .

Let the edge set $E (Γ (Z_{p} \times Z_{p} \times Z_{p} \dots \times Z_{p})) = \{\begin{matrix} V_{i} is adjacent to V_{j}; i + j ⩽ m \\ V_{i} is non-adjacent to V_{j}; i + j > n \end{matrix}$ Hence, the result.

It is obtained from the above results, we can find the following theorem.

Theorem 5.2

For any odd prime p, then $Γ (Z_{p} \times Z_{p} \times Z_{p} \times \dots \times Z_{p})$ is complete decomposible into cycle of length 4.

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

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Introduction

Decomposition of Γ ( Z p × Z p )

Decomposition of Γ ( Z p × Z p × Z p )

Decomposition of Γ ( Z p × Z p × Z p × Z p )

Decomposition of Γ ( Z p × Z p × Z p × … × Z p )

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[1] Abueida A.A., Daven M., . Multidesigns for graph-pairs of order 4 and 5. Graphs Combin.. 2003;19:433-447.
[Google Scholar]

[2] Abueida A.A., Daven M., . Multidecompositions of the complete graph. Ars Combinatoria.. 2004;72:17-22.
[Google Scholar]

[3] Abueida A.A., O’Neil T., . Multidecomposition of $λ K_{m}$ into small cycles and claws. Bull. Inst. Combin. Appl.. 2007;49:32-40.
[Google Scholar]

[4] Adams P., Bryant D.E., Khodkar A., . On Alspach’s conjecture with two even cycle lengths. Discrete Math.. 2000;223:1-12.
[Google Scholar]

[5] Alspach B., Marshall S., . Even cycle decompositions of complete graphs minus a $1$ -factor. J. Comb. Des.. 1994;2:441-458.
[Google Scholar]

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Decomposition of zero divisor graph into cycles and stars

Abstract

Keywords

Graph decomposition

Zero divisor graph

Balanced complete bipartite

05C25

06A11

05C70

1 Introduction

2 Decomposition of Γ ( Z p × Z p )

3 Decomposition of Γ ( Z p × Z p × Z p )

4 Decomposition of Γ ( Z p × Z p × Z p × Z p )

5 Decomposition of Γ ( Z p × Z p × Z p × … × Z p )

Declaration of Competing Interest

References

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4 Decomposition of $Γ (Z_{p} \times Z_{p} \times Z_{p} \times Z_{p})$

5 Decomposition of $Γ (Z_{p} \times Z_{p} \times Z_{p} \times \dots \times Z_{p})$